Question

An hourglass sand timer drips 2 cm^3 of sand every minute. It has a radius of 4cm and a height of 15cm. When there is 7cm of sand in the hourglass, what rate is the depth of the sand decreasing (dv/dt) AND what is the rate that the radius is changing (dr/dt)? (This is a cone problem)

HOW I NEED IT SET UP:

EXAMPLE BUT WITH CYLINDER LENGTH AND RADIUS:

v(t)= π ((r(t)^2)’ l(t) + r(t)^2 * l’(t))

product rule: Dv/dt= π (2r(t)r’(t)*l(t)+r(t)^2 * l’(t)

Derivative: π (2r(t)r’(t)+r(t)^2*l’(t)

Da/dt= 4π r(t) r’ (t) + 2π (r’ (t) l(t) + r(t) l’ (t))

*****AFTER SHOWING THESE STEPS THEN PLUG IN VARIABLES AND SOLVE FOR THE RATE OF VOLUME AND RADIUS CHANGING. THANKS SO MUCH! This is my 3rd time posting this question PLEASE DON'T USE THE PREVIOUS WRONG ANSWERS!******

-INDICATE ALL DERIVATIVES YOU ARE USING AND SHOW ALL STEPS FOR THUMBS UP!!!

Answer 1

kindly please upvote this

a) Geren, ov, -2 cm? /min Hemozione When h=1, r= ? tano= cod Forconst] - htanor - at tano = dro 4 r = 1854 7-28/15 J Nove [Here, h= depth for hh v=forth av rohor o at Sometimes - Zeniyetet 2X1 X3 X7*(*}x} + {X AX en -0.1827 em min & alig on tono = -0.1827 x 4 Home -0.04827 cm/min) se CU